What is the solution for y^{''}+5y^'+4y=1 ?

The solution for y^{''}+5y^'+4y=1 is y=c_{1}e^{-t}+c_{2}e^{-4t}+1/4

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y^{''}+5y^'+4y=1

y^{^{''}} + 5 y^{^{'}} + 4 y = 1

y = c_{1} e^{- t} + c_{2} e^{- 4 t} + \frac{1}{4}

Solution steps

y^{''} (t) + 5 y^{'} (t) + 4 y = 1

Solve linear ODE:

y = c_{1} e^{- t} + c_{2} e^{- 4 t} + \frac{1}{4}

y = c_{1} e^{- t} + c_{2} e^{- 4 t} + \frac{1}{4}

y^{^{''}} + 2 y^{^{'}} = 3 e^{- 2 t}

y^{^{''}} - 4 y = e^{x}

y^{^{''}} - 4 y = - 12 e^{- 2 x} + 15 cos (x) + 8 x

y^{^{''}} + 2 y^{^{'}} + y = \frac{e ^{- x}}{1 + x ^{2}}

y^{^{''}} - 4 y = (x^{2} - 9) sin (2 x)

What is the solution for y^{''}+5y^'+4y=1 ?
The solution for y^{''}+5y^'+4y=1 is y=c_{1}e^{-t}+c_{2}e^{-4t}+1/4